Find N in a Limit of a Sequence Homework

[kwal0203]

Homework Statement

Suppose a_{n}=\frac{n^2-2n+1}{2n^2+4n-1}

For each positive number \epsilon, find a number N such that:

\mid a_{n} - L\mid < \epsilon whenever n > N.

Homework Equations

The Attempt at a Solution

\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon

\mid \frac{3-8n} {4n^2+8n-2} \mid < \epsilon

Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

Thanks!

 

[PeroK]

Homework Statement

Suppose a_{n}=\frac{n^2-2n+1}{2n^2+4n-1}

For each positive number \epsilon, find a number N such that:

\mid a_{n} - L\mid < \epsilon whenever n > N.

Homework Equations

The Attempt at a Solution

\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid < \epsilon

\mid \frac{3-8n} {2n^2+4n-1} \mid < \epsilon

Now I have no idea how to isolate the n so that I can find the N value. Any help appreciated.

Thanks!

Try some estimating.

Can you fix your latex?

 

[kwal0203]

Try some estimating.

Can you fix your latex?

Fixed it up.

Estimating what exactly?

 

[PeroK]

Fixed it up.

Estimating what exactly?

Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

Show that $5n^2 + 3n + 133$ tends to $\infty$ an $n \rightarrow \infty$

Estimating technique:

For $n > 1$ we have $n^2 > n$ hence $5n^2 > n$. So:

$5n^2 + 3n + 133 > 4n + 133 > 4n > n$

Now, we can use a simple $N$ in the limit argument. Whereas, trying to find $N$ for the orginal quadratic would be messy.

 

[PeroK]

\mid \frac{3-8n} {2n^2+4n-1} \mid &lt; \epsilon

PS I think you've lost a factor of $2$ from the denominantor.

 

[kwal0203]

PS I think you've lost a factor of $2$ from the denominantor.

Yes I definitely have, thanks.

 

[kwal0203]

Estimating is a technique, which no one seems to teach explicitly for evaluating limits. Let me give you an example and see if you can apply the idea to your case.

Show that $5n^2 + 3n + 133$ tends to $\infty$ an $n \rightarrow \infty$

Estimating technique:

For $n > 1$ we have $n^2 > n$ hence $5n^2 > n$. So:

$5n^2 + 3n + 133 > 4n + 133 > 4n > n$

Now, we can use a simple $N$ in the limit argument. Whereas, trying to find $N$ for the orginal quadratic would be messy.

So it goes kinda likes this:

For n > 1 we have n^2&gt;n hence, 2n^2&gt;n So:

2n^2+4n-1&gt;5n-1&gt;5n&gt;n

And in a similar way:

For n > 1 we have 3 - 8n &gt; 3 - 8 = -5 So:

\mid \frac{-5}{n} \mid &lt; \epsilon

and,

n &gt; 5 / \epsilon

Is that what you mean?

 

[PeroK]

So it goes kinda likes this:

For n > 1 we have n^2&gt;n hence, 2n^2&gt;n So:

2n^2+4n-1&gt;5n-1&gt;5n&gt;n

And in a similar way:

For n > 1 we have 3 - 8n &gt; 3 - 8 = -5 So:

\mid \frac{-5}{n} \mid &lt; \epsilon

and,

n &gt; 5 / \epsilon

Is that what you mean?

First, you need to take that $\epsilon$ out of your working. That seems to be a common mistake to introduce $\epsilon$ into the inequality too soon. Second, your estimates have gone a bit wonky!

Let me show you what I mean. You should leave out the $\epsilon$ for now and do:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots$

Note that the numerator is less than $8n$ and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because $8n/n$ doesn't converge to 0. So, you need an estimate based on $n^2$ in this case.

It's only when you've finished this algebra that you bring in $\epsilon$. For example:

Let $\epsilon > 0$ and take $N > 1/ \epsilon$ (or whatever you think you need from your calculations). Then $n > N$ implies:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon$

 

[kwal0203]

First, you need to take that $\epsilon$ out of your working. That seems to be a common mistake to introduce $\epsilon$ into the inequality too soon. Second, your estimates have gone a bit wonky!

Let me show you what I mean. You should leave out the $\epsilon$ for now and do:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1} {2} \mid \ = \ \mid \frac{3-8n} {4n^2+8n-2} \mid \ < \ \mid \frac{8n} {4n^2+8n-2} \mid \ \dots$

Note that the numerator is less than $8n$ and you need an over-estimate on the numerator and an underestimate on the denominator. But, your underestimate on the denominator can't be too crude, because $8n/n$ doesn't converge to 0. So, you need an estimate based on $n^2$ in this case.

It's only when you've finished this algebra that you bring in $\epsilon$. For example:

Let $\epsilon > 0$ and take $N > 1/ \epsilon$ (or whatever you think you need from your calculations). Then $n > N$ implies:

$\mid \frac{n^2 -2n + 1} {2n^2+4n-1} - \frac{1}{2} \mid < \frac{1}{n} < \epsilon$

Ok I think I get it now. Thanks for your help I'll do some more work on it.

 

[PeroK]

Ok I think I get it now. Thanks for your help I'll do some more work on it.

The simplest way to explain estimating is perhaps:

$|\frac{a(n)}{b(n)}| < |\frac{c(n)}{d(n)}|$

Where $|a(n)| < |c(n)|$ and $|b(n)| > |d(n)|$ and $c(n), d(n)$ are some simple expressions in $n$ that reduce to something that can easily be shown to have a limit of $0$.